19. A diploid cell has 40 chromosomes. How many chromosomes are present in each cell once the mitotic cycle has concluded, including cytokinesis? A. 10 B. 20 C. 40 D. 80 E. 160

Answers

Answer 1

The correct answer is B:20. A diploid cell has 40 chromosomes. How many chromosomes are present in each cell once the mitotic cycle has concluded, including cytokinesis.

A diploid cell contains two complete sets of chromosomes, so 2n is equal to 40 chromosomes in this scenario. Mitosis is the process of cell division in which two new cells are formed from one original cell, each containing the same number of chromosomes as the original cell. So, after mitosis is completed, each of the two daughter cells will contain the same number of chromosomes as the original cell: 40 chromosomes divided equally between the two daughter cells results in 20 chromosomes in each cell.

The chromosomal number does not vary in a mitotic cell, unlike in meiotic cell division where it gets reduced by half. Mitosis is a process that is used to create new cells in multicellular organisms. In mitosis, a cell splits into two identical daughter cells. A complete set of chromosomes is included in each daughter cell, and the DNA is duplicated in each cell.The correct answer is B.

To know more about chromosomes visit:

https://brainly.com/question/30077641

#SPJ11

Answer 2

A diploid cell having 40 chromosomes, at the end of mitosis and cytokinesis, each daughter cell will have 40 chromosomes. Option (C) 40 is the correct answer.

Mitosis is the process of cell division that produces two identical daughter cells. Cytokinesis is the final step of mitosis that splits the cytoplasm of a cell and causes the formation of two daughter cells at the end of mitosis.

The result of mitosis and cytokinesis is two daughter cells, each with the same amount of genetic material as the parent cell and with the same number of chromosomes.

Therefore, the diploid cell having 40 chromosomes will produce two identical daughter cells at the end of mitosis, each having 40 chromosomes. Hence, the correct option is (C) 40.

To know more about chromosomes, visit:

https://brainly.com/question/30077641

#SPJ11


Related Questions

Which of the following is a correct definition of hyponatremia? O Elevated blood sodium concentration O Low blood sodium concentration

Answers

Hyponatremia is a condition where there is low blood sodium concentrationHyponatremia is a medical condition wherein there is an electrolyte imbalance, and the sodium level in the blood is low. Sodium is an important electrolyte in the body, and its level is closely regulated by the body's water balance.

The condition can occur when the intake of sodium is low, when the body retains more water than it should, or when both conditions occur simultaneously.What are the causes of Hyponatremia?Hyponatremia is primarily caused by the following conditions:

Vomiting and diarrhea: These conditions cause loss of body fluids which could lead to hyponatremia.Sweating: It causes loss of fluids from the body, and if these fluids are not replaced with water and electrolytes, hyponatremia may occur.Certain medications: Some medications like diuretics can cause hyponatremia if taken in large doses and over a long period of time.Adrenal gland failure: It could lead to low levels of sodium in the blood. The adrenal gland produces hormones that regulate the balance of fluids in the body, and its failure could cause an electrolyte imbalance.What are the symptoms of Hyponatremia?The symptoms of hyponatremia depend on the severity of the condition. Mild hyponatremia often does not cause any symptoms, but when the condition becomes severe, symptoms like nausea, headache, confusion, seizures, coma, and death may occur.

TO know more about that Hyponatremia visit:

https://brainly.com/question/28432747

#SPJ11

15) UTI's with microbial etiology include: A. cystitus. B. Urethritis C. Leptospirosis D. A and B E. A, B and C 16) The cause of gonorrhea is a member of the genus: A. Borrelia B. treponema C. Neisseria D. Mycobacterium E. plasmodium 17) Which antibody is most import in immediate hypersensitivity reactions: A. IgG B. IgM C. IgA D. ISE 18) Which is true. Of. HPV (papillomavirus) A. Only two strains. Effect humans B. It can cause genital warts C. Less than 1% of women are effected D. No vaccine is available 19). Trichomonal. Vaginitis is caused by: A. Yeast B. Bacteria C. Protozoan D. Chlamydia E. A virus 20) Lyme disease A. Is highly contagious B. Early symptoms include rash and flu like symptoms etiology D. Mosquito vector C. Viral

Answers

UTIs with microbial etiology include cystitis and urethritis. The cause of gonorrhea is a member of the genus Neisseria. The most important antibody in immediate hypersensitivity reactions is IgE.

UTIs (urinary tract infections) with microbial etiology commonly involve cystitis (inflammation of the bladder) and urethritis (inflammation of the urethra). These infections are often caused by bacterial pathogens.

Gonorrhea is caused by a member of the genus Neisseria, specifically Neisseria gonorrhoeae, a sexually transmitted bacterium.

In immediate hypersensitivity reactions, the most important antibody involved is IgE. IgE antibodies are responsible for triggering allergic reactions and are associated with conditions like asthma and allergic rhinitis.

HPV (human papillomavirus) is a sexually transmitted infection that can cause genital warts and is also associated with certain types of cancer. There are several strains of HPV that affect humans, not just two, and there is a vaccine available to protect against certain high-risk strains.

Trichomonal vaginitis, also known as trichomoniasis, is caused by a protozoan parasite called Trichomonas vaginalis.

Lyme disease is primarily transmitted through the bite of infected black-legged ticks. It is not highly contagious between humans. Early symptoms of Lyme disease often include a characteristic rash called erythema migrans, along with flu-like symptoms.

Learn more about Gonorrhea here:

https://brainly.com/question/32219831

#SPJ11

State one possible hypothesis that can explain the global distribution of lactase persistence (lactose tolerance) and lactase nonpersistance (lactose intolerance). Be sure to include the following keywords in your explanation; selection, fitness, survival.

Answers

The natural selection, fitness hypothesis suggests the global distribution of lactase persistence and non persistence may have arisen an adaptive response to availability or absence of dairy farming practices.

One possible hypothesis to explain the global distribution of lactase persistence (lactose tolerance) and lactase nonpersistence (lactose intolerance) is the "natural selection and fitness" hypothesis. This hypothesis suggests that lactase persistence may have been positively selected for in populations that traditionally relied on dairy consumption as a significant source of nutrients, while lactase non persistence may have been advantageous in populations with limited or no history of dairy farming.

In regions where dairy farming has been prevalent for thousands of years, individuals with the genetic mutation that allows for lactase persistence would have had a survival advantage. The ability to digest lactose, the sugar present in milk, would have provided a valuable source of nutrition, especially during times of scarcity or limited food resources. This increased fitness and survival among lactase-persistent individuals would have led to a higher prevalence of the lactase persistence trait in these populations over generations.

Learn more about lactase persistence here

https://brainly.com/question/32484975

#SPJ11

In photosynthesis, carbon dioxide is "fixed" in ____.
A. the light-dependent reactions
B. the Carbon cycle
C. the light-independent reactions
D. the Krebs cycle

Answers

The correct answer is C. the light-independent reactions, where carbon dioxide is fixed and converted into organic compounds during photosynthesis.

The process of carbon dioxide fixation refers to the conversion of atmospheric carbon dioxide into organic compounds during photosynthesis. This occurs during the light-independent reactions, also known as the Calvin cycle or the dark reactions. These reactions take place in the stroma of chloroplasts, specifically in the chloroplasts of plant cells.

During the light-independent reactions, carbon dioxide molecules are combined with molecules derived from the light-dependent reactions, such as ATP and NADPH. The key enzyme involved in carbon dioxide fixation is called RuBisCO (Ribulose-1,5-bisphosphate carboxylase/oxygenase). RuBisCO catalyzes the incorporation of carbon dioxide into an organic molecule called ribulose-1,5-bisphosphate (RuBP), which then goes through a series of reactions to produce glucose and other organic compounds.

In contrast, the light-dependent reactions, which occur in the thylakoid membranes of chloroplasts, involve the absorption of light energy and the generation of ATP and NADPH. These energy-rich molecules produced in the light-dependent reactions are subsequently used in the light-independent reactions to drive the carbon dioxide fixation and synthesis of organic molecules.

Learn more about photosynthesis here:

https://brainly.com/question/29764662

#SPJ11

How might your immune system use MHC II to eliminate a viral
invader? How is this different from using MHC I?

Answers

The immune system employs MHC II molecules to eliminate viral invaders. MHC II differs from MHC I in terms of the antigen presentation pathway it employs.

The immune system utilizes Major Histocompatibility Complex (MHC) molecules to detect and present antigens to immune cells. MHC II molecules are primarily found on the surface of antigen-presenting cells, such as dendritic cells, macrophages, and B cells.

When a viral invader enters the body, antigen-presenting cells engulf the virus and break it down into smaller protein fragments. These protein fragments, known as antigens, are then loaded onto MHC II molecules within the antigen-presenting cells.

The MHC II molecules with the viral antigens are then transported to the cell surface and presented to CD4+ T cells, which recognize and bind to the antigen-MHC II complex. This interaction activates the CD4+ T cells, enabling them to coordinate an immune response to eliminate the viral invader. The MHC II pathway is critical for activating helper T cells and initiating an adaptive immune response against viral infections.

In contrast, MHC I molecules are found on the surface of almost all nucleated cells in the body. They are responsible for presenting antigens derived from intracellular proteins, including viral proteins synthesized within infected cells. Infected cells process viral proteins into antigenic peptides, which are then loaded onto MHC I molecules.

The MHC I-antigen complex is presented on the cell surface, where it is recognized by CD8+ T cells. This recognition triggers the destruction of the infected cells by cytotoxic T cells, preventing the virus from spreading further. The MHC I pathway is crucial for identifying and eliminating virus-infected cells.

Learn more about MHC II molecules here:

https://brainly.com/question/31592689

#SPJ11

2. Enterobius vermicularis is infective in___ form and causes ____
a. larval; pinworm
b. egg; hookworm
c. egg; pinworm d.larval; hookworm 3. The reproductive structure of Taenia is a a.hook b.proglottid c. scolex d.heterocyst
4. Trichinella spiralis is transmitted by
a. ingestion of a cyst b. ingestion of a larva
c. ingestion of an egg d.a vector 5. Which type of sample would be used to aid in diagnosis of a Clonorchis infection? a. Fecal smear b.Sputum sample
c. Skin scraping d.Blood sample

Answers

Enterobius vermicularis is infective in the egg form and causes pinworm infection. The reproductive structure of Taenia is the proglottid. Trichinella spiralis is transmitted by ingestion of a larva. A fecal smear would be used to aid in the diagnosis of a Clonorchis infection.

Enterobius vermicularis is infective in the egg form and causes pinworm infection. The eggs of Enterobius vermicularis are ingested, usually through contaminated food, water, or by direct contact with infected individuals. Once inside the body, the eggs hatch in the small intestine, and the larvae migrate to the large intestine, where they mature into adult worms. The adult female worms then migrate to the perianal area to lay their eggs, leading to itching and discomfort.

The reproductive structure of Taenia, a genus of parasitic tapeworms, is the proglottid. Proglottids are segments that make up the body of a tapeworm and contain both male and female reproductive organs. Each proglottid is capable of producing eggs, which are then released into the environment through the feces of the infected host. The proglottids can detach from the tapeworm's body and be passed in the feces, enabling the tapeworm to spread and infect new hosts.

Trichinella spiralis, a parasitic roundworm, is transmitted by the ingestion of a larva. The larvae of Trichinella are encysted in the muscle tissue of infected animals, typically pigs or other mammals. When these infected meat products are consumed by humans, the larvae are released in the digestive system, where they mature into adult worms. The female worms then produce larvae that migrate to muscle tissue, causing a condition known as trichinellosis.

To aid in the diagnosis of a Clonorchis infection, a fecal smear would be used. Clonorchis sinensis is a parasitic liver fluke that infects humans through the consumption of raw or undercooked freshwater fish containing the infectious larvae. The adult flukes reside in the bile ducts of the liver. The presence of Clonorchis eggs in a fecal smear can indicate an infection, as the adult flukes release eggs into the feces. Other diagnostic methods may include serological tests or imaging techniques to visualize the flukes in the bile ducts.

Know more about Trichinella spiralis here:

https://brainly.com/question/30978492

#SPJ11

150 words each
1.. Would you expect cells in different organs in complex animals to have the same structures? Explain your reasoning and give examples.
2. How do cell junctions help to form tissues? Compare two tissues of your choice and explain the benefits of junctions in those tissues.

Answers

In both tissues, cell junctions facilitate the formation of cohesive and functional tissues. They provide mechanical strength, enable cell communication, and allow tissues to perform

1. No, cells in different organs in complex animals would not be expected to have the same structures. The reason for this is that different organs have specialized functions and require specific structures to perform their respective roles effectively, For example, the cells in the heart muscle (cardiomyocytes) have unique structures such as intercalated discs, which allow for synchronized contraction of the heart muscle.

In contrast, cells in the epithelial lining of the small intestine have structures called tight junctions. These junctions form a barrier between adjacent cells, preventing the leakage of digested nutrients and other molecules from the intestinal lumen into the underlying tissues. The tight junctions in the intestinal epithelium help in maintaining the integrity of the digestive system and facilitating the absorption of nutrients.

2. Cell junctions play a vital role in forming tissues by providing structural integrity, communication, and coordination among cells. Two tissues where cell junctions are essential are epithelial tissue and cardiac muscle tissue. In epithelial tissue, tight junctions play a crucial role in forming a barrier between cells, preventing the passage of substances between cells.

In cardiac muscle tissue, intercalated discs containing gap junctions are critical for the synchronized contraction of the heart. Gap junctions allow the passage of ions and electrical signals between adjacent cardiac muscle cells, enabling rapid and coordinated depolarization and contraction. This ensures that the heart functions as a single unit, efficiently pumping blood throughout the body.

Learn more about functional tissues here

https://brainly.com/question/29159446

#SPJ11

Which one of the following measurements represents a
greater diagnostic value for assessing conditions such as COPD?
a)Flow rate b)Total lung volume. c)Total lung capacity d)Tidal
volume
In the tidal

Answers

Option a is correct. The measurement that represents greater diagnostic value for assessing conditions such as COPD is the flow rate.

When evaluating conditions like COPD, the flow rate is a crucial measurement for diagnostic purposes. Flow rate refers to the speed at which air moves in and out of the lungs during breathing. In COPD, the airways become narrowed and obstructed, leading to difficulty in exhaling air.

By measuring the flow rate, healthcare professionals can assess the severity of airway obstruction and monitor the progression of the condition. On the other hand, while measurements like total lung volume, total lung capacity, and tidal volume provide important information about lung function, they may not directly reflect the degree of airway obstruction characteristic of COPD.

Therefore, the flow rate is considered a more specific and valuable measurement for diagnosing and managing COPD.

Learn more about total lung capacity here:

https://brainly.com/question/31230173

#SPJ11

Complete the Punnet Square and give the phenotype and Genotype: AaBbCe (mom) AABBcc (dad) A- Tall; aa = short B = fat; bb is skinny C = ugly; cc = gorgeous Mom must go on the top.

Answers

Possible phenotypes and genotypes from the cross are: Tall, fat, and ugly (AABBCc), Tall, fat, and attractive (AABbCc), Short, fat, and ugly (AaBBCc), Short, fat, and attractive (AaBbCc).

To complete the Punnett square, we will consider the inheritance of three traits: height (A/a), body shape (B/b), and attractiveness (C/c). Here's the Punnett square:

```

      Aa                Bb              Cc

AABBCc   |   AABBcc   |   AaBBcc

AABbCc   |   AABbcc   |   AaBbcc

AABBCc   |   AABBcc   |   AaBBcc

AABbCc   |   AABbcc   |   AaBbcc

```

Phenotypes and Genotypes:

1. AABBcc: Tall, fat, and ugly (Genotype: AABBCc)

2. AABbcc: Tall, fat, and attractive (Genotype: AABbCc)

3. AaBBcc: Short, fat, and ugly (Genotype: AaBBCc)

4. AaBbcc: Short, fat, and attractive (Genotype: AaBbCc)

So, the possible phenotypes and genotypes from the cross between the mom (AaBbCe) and dad (AABBcc) are:

- Tall, fat, and ugly (AABBCc)

- Tall, fat, and attractive (AABbCc)

- Short, fat, and ugly (AaBBCc)

- Short, fat, and attractive (AaBbCc)

Learn more about Punnett Square at https://brainly.com/question/3522181

#SPJ11

Hypothetical gene "stress-free1" (STF1) is transcriptionally inactive unless cortisol is present.
In addition to DNA elements in the core promoter, there are also silencer elements and enhancer elements. Briefly explain how each silencers and enhancers contribute to the regulation of gene transcription in general then propose a model for how each of these elements might function to ensure that transcription of STF1 is actively expressed only when cortisol is present.

Answers

Silencers and enhancers are DNA elements located upstream of the gene's core promoter and contribute to the regulation of gene transcription in general. Silencers are regions of DNA that bind to transcription factors, preventing the binding of RNA polymerase to the promoter region, thereby reducing or blocking the transcription of the gene.

On the other hand, enhancers are DNA sequences that bind to transcription factors, which increases the likelihood of RNA polymerase binding to the promoter, enhancing gene expression. Gene regulation by enhancers and silencers is usually tissue-specific, depending on the availability of various transcription factors and other regulatory proteins.To ensure that transcription of STF1 is only activated when cortisol is present, the silencer and enhancer elements may function as follows:

Enhancer elements: Cortisol binds to a receptor located upstream of the enhancer region, leading to a conformational change that enables the receptor to bind to the enhancer element. The enhancer element then binds to transcription factors, which leads to RNA polymerase's recruitment, enhancing transcription of STF1.

Silencer elements: In the presence of cortisol, a repressor binds to a DNA element located upstream of the STF1 gene's promoter region, preventing the binding of RNA polymerase, leading to the suppression of transcription.

In the absence of cortisol, the repressor element is inactivated, and the promoter region is free to bind RNA polymerase, leading to transcription of the STF1 gene.

To know more about DNA visit:

https://brainly.com/question/30993611

#SPJ11

How does the occipital lobe interact with the temporal, frontal
and parietal lobe in the formation of an image or visual
system?

Answers

The occipital lobe plays a crucial role in the formation of an image or visual system by processing visual information received by the eyes and sending it to the thalamus. The thalamus then sends the information to the temporal, frontal, and parietal lobes where it is further processed to form an image or visual system.

The occipital lobe interacts with the temporal, frontal, and parietal lobe in the formation of an image or visual system through the optic nerve and the thalamus.

The occipital lobe is responsible for processing visual information. The temporal, frontal, and parietal lobes are also involved in the processing of visual information but they play different roles in the formation of an image or visual system.

The temporal lobe is responsible for recognizing objects, faces, and colors. It is involved in the identification of objects and the recognition of faces.

The frontal lobe is involved in the processing of visual information related to motion, depth perception, and spatial awareness. It plays an important role in the formation of an image or visual system.The parietal lobe is responsible for the integration of information from the different sensory systems.

It processes information related to spatial awareness and the orientation of the body in space.The visual information received by the eyes is sent to the occipital lobe through the optic nerve. The occipital lobe processes this information and sends it to the thalamus.

The thalamus then sends the information to the temporal, frontal, and parietal lobes where it is further processed to form an image or visual system.

In summary, the occipital lobe plays a crucial role in the formation of an image or visual system by processing visual information received by the eyes and sending it to the thalamus. The thalamus then sends the information to the temporal, frontal, and parietal lobes where it is further processed to form an image or visual system.

To know more about occipital visit;

brainly.com/question/30767795

#SPJ11

what is the total amount of ATP generated in cellular
respiration?
please help quickly with very very short explination!

Answers

A total of up to 36 molecules of ATP can be consisted from just one molecule of glucose in the work of cellular respiration.

What is the ATP?

The total amount of ATP create in basic breathing changes depending on the particular road complicated. Electron transport from the particles of NADH and FADH2 from glycolysis, the revolution of pyruvate, and the Krebs cycle generates as many as 32 more ATP particles.

Therefore,  In general, through the complete disintegration of individual particle of hydrogen, the net result of ATP is 36 to 38 particles in prokaryotes and 30 to 32 fragments in eukaryotes.

Learn more about ATP  from

https://brainly.com/question/893601

#SPJ4

Achondroplasia is caused by mutations in the Fibroblast growth factor receptor 3 gene. It is a disorder of bone growth that prevents the changing of cartilage to bone. O Statement 1 is correct. Statement 2 is incorrect Both statements are incorrect Statement 1 is incorrect. Statement 1 is correct. Both statements are correct Neurofibromatosis 1 is considered an autosomal dominant disorder because the gene is located on the long arm of chromosome 17. It is caused by microdeletion at the long arm of chromosome 17 band 11 sub-band 2 involving the NF1 gene. Both statements are incorrect O Both statements are correct O Statement 1 is correct. Statement 2 is incorrect O Statement 1 is incorrect, statement 2 is correct Genetic disorder is a disease that is caused by an abnormality in an individual's DNA. Range from a small mutation in DNA or addition or subtraction of an entire chromosome or set of chromosomes. O Both statements are correct Statement 1 is correct. Statement 2 is incorrect O Statement 1 is incorrect, statement 2 is correct O Both statements are incorrect.

Answers

The correct option is "Statement 1 is correct, Statement 2 is incorrect."Genetic disorders are diseases caused by abnormalities in an individual's DNA.

They can range from a small mutation in DNA to the addition or subtraction of an entire chromosome or set of chromosomes.Achondroplasia is a disorder of bone growth that prevents the changing of cartilage to bone. It is caused by mutations in the Fibroblast growth factor receptor 3 gene.

Statement 1 is correct about Achondroplasia.Neurofibromatosis 1 is caused by microdeletion at the long arm of chromosome 17 band 11 sub-band 2 involving the NF1 gene. Neurofibromatosis 1 is considered an autosomal dominant disorder because the gene is located on the long arm of chromosome 17. Statement 2 is incorrect about Neurofibromatosis 1.

To know more about diseases visit:

https://brainly.com/question/17950058

#SPJ11

Why taxonomic nomenclature is important? It provides the unified language for communication about biological diversity. It reflects evolutionary relatedness of taxa. Scientific names often capture important characteristics of the animals. It documents the history of science. All of the above.

Answers

Taxonomic nomenclature is important because it provides a standardized language for communication, represents evolutionary relationships, captures important characteristics, and documents the history of scientific discoveries. So, All of the above is the correct choice.

Taxonomic nomenclature is important for several reasons:

It provides a unified language for communication about biological diversity: By assigning unique scientific names to organisms, taxonomic nomenclature allows researchers, scientists, and other professionals to communicate and exchange information accurately and precisely. This ensures clarity and avoids confusion that may arise from using different common names for the same species.It reflects evolutionary relatedness of taxa: Taxonomic nomenclature is based on the principles of evolutionary relationships. Organisms with similar characteristics and shared ancestry are grouped together into taxa (such as genus, family, order, etc.), and their scientific names reflect their evolutionary relationships. This helps in understanding the evolutionary history and biological relationships between different organisms.Scientific names often capture important characteristics of the animals: Scientific names are often chosen to describe important characteristics of the organisms they represent. These names can provide insights into the morphology, behavior, habitat, or other significant features of the species. This additional information enhances our understanding of the organism beyond its common name.It documents the history of science: Taxonomic nomenclature has a long history and has evolved over time. The use of scientific names allows us to trace the development of scientific knowledge, discoveries, and advancements in the field of taxonomy. The history of taxonomic naming provides valuable insights into the progression of scientific understanding and serves as a record of scientific exploration.

To know more about Taxonomic nomenclature

brainly.com/question/12074909

#SPJ11

Acetyl-CoA is an important intermediate that participates (either as an input, an output, or an intermediate) in all of the below processes EXCEPT O Photorespiration O the Citric Acid Cycle B-oxidation cycle Acetyl-CoA participates in all these processes O Glyoxylate cycle Determination of an enzyme or pathway Q10 provides information on O a method to compare two alternative enzymes or pathways at a single temperature O gas solubility in response to temperature O the relative thermal motivation of a biochemical pathway a O the temperature sensitivity of an enzyme or pathway O the temperature switch point between C3 and CAM photosynthesis

Answers

Acetyl-CoA is an important intermediate that participates in all of the processes mentioned except gas solubility in response to temperature.

Option (F) is correct.

Acetyl-CoA is a central molecule in cellular metabolism. It is involved in various biochemical processes, including the ones mentioned:

A) Photorespiration: Acetyl-CoA participates in photorespiration as an input in the glycolate pathway, which helps plants recover carbon during inefficient photosynthesis.

B) The Citric Acid Cycle: Acetyl-CoA enters the citric acid cycle, also known as the Krebs cycle, where it undergoes a series of reactions to generate energy-rich molecules such as ATP.

C) β-oxidation cycle: Acetyl-CoA is produced as an output during the breakdown of fatty acids in the β-oxidation cycle, which occurs in mitochondria.

D) Glyoxylate cycle: Acetyl-CoA serves as an intermediate in the glyoxylate cycle, allowing certain microorganisms and plants to convert acetyl-CoA into carbohydrates.

E) Determination of an enzyme or pathway Q10: Acetyl-CoA can participate in the determination of the temperature sensitivity of an enzyme or pathway using the Q10 value, which describes the rate of change with temperature.

However, F) Gas solubility in response to temperature does not involve Acetyl-CoA directly. It refers to the solubility of gases, such as oxygen or carbon dioxide, in liquids and is influenced by factors like temperature and pressure.

To learn more about Acetyl-CoA  here

https://brainly.com/question/16000193

#SPJ4

Complete question is:

Acetyl-CoA is an important intermediate that participates (either as an input, an output, or an intermediate) in all of the below processes EXCEPT:

A) Photorespiration

B) The Citric Acid Cycle

C) β-oxidation cycle

D) Glyoxylate cycle

E) Determination of an enzyme or pathway Q10 provides information on

F) Gas solubility in response to temperature

G) The relative thermal motivation of a biochemical pathway

H) The temperature sensitivity of an enzyme or pathway

I) The temperature switch point between C3 and CAM photosynthesis

Urine with fixed specific gravity is a distinctive feature of acute renal failure. Select one: True False

Answers

False, urine with fixed specific gravity is not a distinctive feature of acute renal failure.

Explanation:Urine with a fixed specific gravity is when the kidney is unable to concentrate or dilute urine in response to changes in water intake.

The specific gravity of urine can be used to detect kidney disease or injury.

In acute renal failure, the kidneys are unable to filter waste products from the blood effectively, resulting in an accumulation of toxins in the bloodstream.

This leads to a variety of symptoms and may be caused by a number of factors including injury, infection, or medication.

A decrease in urine output or anuria, a significant increase in blood pressure, electrolyte imbalances, and accumulation of nitrogenous waste products in the blood can all be signs of acute renal failure. Urine with a fixed specific gravity is not a distinctive feature of acute renal failure.

Therefore, the statement "Urine with fixed specific gravity is a distinctive feature of acute renal failure" is false.

To know more about urine visit;

brainly.com/question/31923882

#SPJ11

Chose the correct order of entities according to mutation rate (from lowest to highest, i.e. least mutable to most mutable)? O Viroids, ssRNA viruses, dsDNA viruses, bacteria, eukaryotes Protists, bac

Answers

The correct order of entities according to mutation rate (from lowest to highest, i.e. least mutable to most mutable) is Viroids, ssRNA viruses, dsDNA viruses, bacteria, eukaryotes, Protists, bac.

Biological entities are prone to changes in genetic material from time to time, this change is known as mutations, which is a basic phenomenon of evolution. The speed of mutation varies between biological entities.Viroids have the least mutation rate as they do not encode proteins. They only produce a few gene products that mainly depend on the host's metabolism. ssRNA viruses are a bit more mutable than viroids as RNA is not as stable as DNA, which means errors are more likely to occur during replication. DsDNA viruses are more mutable than RNA viruses as they have an error-correction mechanism that allows them to repair most replication errors.

Bacteria are more mutable than dsDNA viruses as they undergo horizontal gene transfer and have fewer DNA repair mechanisms. Eukaryotes are more mutable than bacteria as they have slower replication and DNA repair mechanisms. Protists are more mutable than eukaryotes as they are unicellular and have high mutation rates. Bacteria, on the other hand, have a high mutation rate because they reproduce rapidly and have horizontal gene transfer that allows them to acquire new genes and share them. So therefore The correct order of entities according to mutation rate (from lowest to highest, i.e. least mutable to most mutable) is Viroids, ssRNA viruses, dsDNA viruses, bacteria, eukaryotes, Protists, bac.

Learn more about eukaryotes at:

https://brainly.com/question/1286526

#SPJ11

Which one is the correct hierarchical sequence of the auditory stimulus processing? (Some intermediate structures may be omitted.)
a) Vesibulocochlear nerve - Inferior Colliculus - Cochlear Nuclei - Medial Geniculate nucleus - Primary Auditory cortex.
b) Cranial nerve VIII - Cochlear Nuclei – Medial Geniculate nucleus - Inferior Colliculus - Primary Auditory cortex.
c) Cranial nerve V - Cochlear Nuclei – Inferior Colliculus - Medial Geniculate nucleus - Primary Auditory cortex.
d) Hair cells – Spiral ganglion cells – Cochlear Nuclei – Inferior Colliculus - Medial Geniculate nucleus - Primary Auditory cortex.

Answers

The correct hierarchical sequence of the auditory stimulus processing is (b) Cranial nerve VIII - Cochlear Nuclei – Medial Geniculate nucleus - Inferior Colliculus - Primary Auditory cortex. Here is an explanation for each of the structures:

Auditory stimulus processing is the step-by-step process that sound waves undergo as they travel from the ear to the brain for interpretation. The structures involved in this process are as follows:

Cranial nerve VIII (CN VIII) or Vestibulocochlear nerve: This is the nerve responsible for transmitting sound information from the ear to the brain.

Cochlear Nuclei: These are two small clusters of cells located in the brainstem. They receive and process sound information from the cochlea.

Medial Geniculate Nucleus: This is a group of nuclei in the thalamus that act as the main relay center for auditory information processing.

Inferior Colliculus: This is a midbrain structure that receives and integrates auditory information from both ears.

Primary Auditory Cortex: This is the first cortical region in the temporal lobe responsible for processing auditory information from the thalamus.

The correct sequence, therefore, is Cranial nerve VIII - Cochlear Nuclei – Medial Geniculate nucleus - Inferior Colliculus - Primary Auditory cortex.

To know more about stimulus visit:

https://brainly.com/question/298785

#SPJ11

Scenario Mr. Johnson is a 70-year-old male complaining of shortness of breath for the past three weeks. Mr. Johnson is complaining that he has chest pain, and this pain increases when he coughs. He also reports thick green/yellow sputum for the past week. His current weight was stable at 100 kg from his previous visit six months ago. He admits to occasionally smoking cigarettes. Mr. Johnson's assessment is as follows: . Inspection upper respiratory system: Nasal and mouth mucosa is pink; no bleeding, masses, or deformities are noted in the upper respiratory system. Inspection lower respiratory system: The client has a respiratory rate of 20 with even and unlabored respirations. During the history, the client is speaking freely and does not report any shortness of breath while talking. • The client has skin appropriate for his ethnic background, with no skin integrity issues noted during the inspection. Palpation: No masses, deformities, or crepitus are noted. Trachea is midline and nontender. . The client has equal lung expansion anterior and posterior; the client reports pain that increases with inspiration. • Percussion: Dullness over right lower lobe, otherwise hyper resonance. . Auscultation: Fine crackles in the right lower lobe with inspiration and expiratory wheezes and diminished breath sounds noted throughout. • Vital signs: Temperature: 100°F (38°C); Respiratory rate: 22; Pulse oximetry on room air: 91% to 93%; Heart rate: 90 bpm; and Blood pressure: 130/80 mm Hg As the nurse, you have determined the priority problem is impaired gas exchange related to the mucus collection in the airways, as evidenced by fine crackles in the right lower lobe. Instructions Using the assessment and nursing diagnosis provided in the scenario, write 200-250 words identifying goals for Mr. Johnson in your initial post. Then, respond to at least two of your peers' posts. Discussion Prompts . Identify two measurable short-term goals for Mr. Johnson. Explain why you chose these goals. . Consider what possible outcomes would change the priority problem. . Define one of these possible outcomes and explain how (and why) it would change the priority problem. Then, identify at least one new measurable goal related to the newly identified problem.

Answers

One new measurable goal related to the newly identified problem of improved lung function is Mr. Johnson will have clear breath sounds in all lung fields on auscultation within 48 hours of treatment. This goal is measurable and would indicate improved gas exchange and lung function.

Two measurable short-term goals for Mr. Johnson include:Goal 1: Mr. Johnson will maintain an oxygen saturation level of greater than 92% on room air as evidenced by pulse oximetry every 4 hours.Goal 2: Mr. Johnson will expectorate thick green/yellow sputum within 24 hours of treatment.In order to improve gas exchange, increasing the oxygen saturation level is essential. By maintaining an oxygen saturation level of greater than 92% on room air, it will help improve Mr. Johnson's breathing and decrease his shortness of breath. This goal is realistic and measurable through pulse oximetry. Another important goal is for Mr. Johnson to expectorate thick green/yellow sputum within 24 hours of treatment. This will decrease the amount of mucus and help clear the airways, which in turn will improve gas exchange.The possible outcome that could change the priority problem is improved lung function. Improved lung function would indicate better gas exchange and increased oxygenation. This could be measured through increased oxygen saturation levels, improved breath sounds on auscultation, and decreased respiratory rate. Improved lung function would change the priority problem by decreasing the risk of hypoxemia and respiratory distress.One new measurable goal related to the newly identified problem of improved lung function is Mr. Johnson will have clear breath sounds in all lung fields on auscultation within 48 hours of treatment. This goal is measurable and would indicate improved gas exchange and lung function.

To know more about auscultation visit:

https://brainly.com/question/30386122

#SPJ11

Some genetic diseases have multiple alleles. If there is a mutation in just one allele, this can result in an individual with the disease. True or False

Answers

The statement is True. Some genetic diseases have multiple alleles. If there is a mutation in just one allele, this can result in an individual with the disease.

The most common form of genetic inheritance is caused by a pair of alleles at the same location on a chromosome. There are, however, multiple variants, called multiple alleles, in some situations. The ABO blood group, for example, is governed by three alleles: A, B, and O. As a result, if an individual has a mutation in only one allele, the disease may be present. Because of the potential for two or more dominant alleles to occur, multiple alleles can lead to different phenotypic outcomes.

An allele is a variant of a gene that is located at a specific point on a chromosome and that determines one or more traits. The term “multiple alleles” refers to the existence of three or more different alleles at the same genetic position. The presence of more than two different alleles at the same locus is referred to as multiple allelism.

To know more about genetic visit:-

https://brainly.com/question/30459739

#SPJ11

"True/False Organismal complexity (how complex an organism is)
is not correlated with genome length but is
correlated with the number of protein coding genes
Group of answer choices
a.True
b.False"

Answers

b. False

Organismal complexity is generally correlated with genome length and not necessarily with the number of protein-coding genes alone. While the number of protein-coding genes contributes to an organism's complexity, it is not the sole determining factor.

Genome length encompasses protein-coding genes and non-coding regions, regulatory elements, repetitive sequences, and other genetic components that contribute to the overall complexity of an organism. Therefore, genome length is a more comprehensive measure of organismal complexity than just the number of protein-coding genes.

Learn more about Organismal complexity-

https://brainly.com/question/15226754

#SPJ11

Numbers 43 through 50 are fill in the blanks. Please be careful and precise. Place these answers in the spaces provide Make sure to note the number of words needed for each. (1 point each/8 total/35) the result of the action of thylako Cidthrin This type of endocytosis is associated with the molecule production of transmembrane integral proteins that can bind to a ligand molecule outside the cell. These proteins are responsible for binding to the ligand and the coating proteins (clathrins) which results in the production of endocytotic vesicle. This form of endocytosis is called endocytosis (two words). toing that are

Answers

The result of the action of thylakoid Clathrins is endocytosis.This type of endocytosis is associated with the molecule production of transmembrane integral proteins that can bind to a ligand molecule outside the cell.

These proteins are responsible for binding to the ligand and the coating proteins (clathrins) which results in the production of endocytotic vesicle. This form of endocytosis is called receptor-mediated endocytosis. Endocytosis is the process of absorbing substances into the cell by enclosing them in a membrane-bound vesicle. It is of three types: Pinocytosis, Phagocytosis, Receptor-mediated endocytosis.

Learn more about  endocytosis:

brainly.com/question/731513

#SPJ11

Pericardial effusion: Please define and describe this diagnosis.
Please name 4 possible causes for this diagnosis. 1-2
paragraph.

Answers

Pericardial effusion is the accumulation of fluid around the heart, which can impair its functioning. It can be caused by factors such as inflammation, heart attack, cancer, and kidney failure.

Pericardial effusion refers to the accumulation of fluid in the pericardial sac, the double-layered membrane that surrounds the heart. It can exert pressure on the heart, impairing its ability to pump blood effectively.

Pericardial effusion can be caused by various factors. Four possible causes include:

Inflammation: Inflammation of the pericardium, known as pericarditis, can lead to pericardial effusion. It may occur due to viral or bacterial infections, autoimmune disorders, or certain medications.

Heart attack: Myocardial infarction (heart attack) can cause damage to the heart muscle, leading to pericardial effusion.

Cancer: Certain types of cancer, such as lung cancer or breast cancer, can metastasize to the pericardium and result in fluid accumulation.

Kidney failure: In some cases, kidney failure can cause an imbalance in fluid levels, leading to pericardial effusion.

To know more about Pericardial effusion, refer to the link:

https://brainly.com/question/32253561#

#SPJ11

Q4: If plants in your home garden displayed a Nitrate deficiency
how would you alleviate the symptoms? (2 marks)

Answers

Nitrate deficiency in plants is caused by the lack of nitrates in the soil. Nitrates are an essential nutrient for plant growth and are responsible for the development of green foliage in plants. If plants in your home garden display a nitrate deficiency, there are several ways to alleviate the symptoms and improve plant growth.

Firstly, the soil should be tested to determine the nitrate level. If the soil is low in nitrate, then it is important to add a fertilizer containing nitrogen. Nitrogen is the main component of nitrates and can be found in fertilizers such as ammonium nitrate or urea. Secondly, adding compost or manure to the soil can also increase the nitrate level.

Lastly, planting leguminous crops such as peas or beans can help to fix nitrogen in the soil, increasing the nitrate level. These methods will help alleviate the symptoms of nitrate deficiency and promote healthy plant growth. The application of fertilizers, compost, manure, and leguminous crops should be done in the right proportions to avoid overuse or underuse of these supplements.

To know more about symptoms visit:-

https://brainly.com/question/29628193

#SPJ11

Which is a main blocking antibody in Immunologic Intervention for Type-I hypersensitivity reaction (desensitization method)? Selected Answer: IgE Answers: IgE IgA IgG IgD IgM .

Answers

The correct answer os IgE.

IgE is the main blocking antibody involved in immunologic intervention for Type-I hypersensitivity reactions during desensitization methods. IgE antibodies are responsible for triggering allergic reactions by binding to allergens and activating mast cells and basophils. Desensitization aims to reduce the hypersensitivity by gradually exposing the individual to increasing doses of the allergen, leading to the production of blocking IgG antibodies that compete with IgE for binding to the allergen, thereby preventing allergic reactions.

In Type-I hypersensitivity reactions, the immune system responds to harmless substances, called allergens, by producing an excessive amount of IgE antibodies. These IgE antibodies bind to the surface of mast cells and basophils, which are rich in histamine. When the individual is re-exposed to the allergen, the allergen binds to the IgE antibodies on the mast cells and basophils, triggering the release of histamine and other inflammatory mediators. This process leads to the symptoms of an allergic reaction, such as itching, swelling, and respiratory difficulties.

To know more about IgE antibodies click here,

https://brainly.com/question/31237327

#SPJ11

The following are stages of glucose oxidation, except. O electron transport system oxidative phosphorylation O Krebs cycle O glycolysis O all of the

Answers

Glucose oxidation is the metabolic process by which glucose is oxidized to produce ATP energy that can be used by the cells for carrying out their activities.

The process of glucose oxidation takes place in three stages, namely glycolysis, Krebs cycle, and electron transport system, which are discussed below.

Glycolysis:

It is the first stage of glucose oxidation that takes place in the cytoplasm of the cell.

In this process, one glucose molecule is oxidized to form two molecules of pyruvic acid.

Moreover, two molecules of ATP energy are produced in this process.

This process can take place in both aerobic and anaerobic conditions.

Krebs Cycle:

It is the second stage of glucose oxidation, also known as the citric acid cycle.

In this stage, the two molecules of pyruvic acid produced during glycolysis are further oxidized to produce energy.

To know more about metabolic visit:

https://brainly.com/question/15464346

#SPJ11

please help
19. Which of the following is the last step that produces inspiration? a. The intrapleural pressure becomes positive b. The diaphragm contracts c. The intercostal muscles contract d. The intra-alveola

Answers

The last step that produces inspiration is that b, the diaphragm contracts.

What is the diaphragm?

The diaphragm is a dome-shaped muscle that separates the chest cavity from the abdominal cavity. When the diaphragm contracts, it flattens and moves down, which increases the volume of the chest cavity. This decrease in intrapleural pressure causes the lungs to expand, which increases the intra-alveolar pressure. This pressure difference causes air to flow into the lungs.

The intercostal muscles are a group of muscles that attach to the ribs. When these muscles contract, they pull the ribs up and out, which also increases the volume of the chest cavity. This increase in volume causes the lungs to expand and air to flow into them.

The intra-alveolar pressure is the pressure inside the alveoli, which are the tiny sacs in the lungs where gas exchange takes place. The intra-alveolar pressure decreases during inspiration, which causes air to flow into the alveoli.

Find out more on inspiration here: https://brainly.com/question/29754321

#SPJ1

What provides the energy to the ATP Synthase for the formation of ATP? (Select all that apply) a. Proton Flow b. Electron Flow c. Phosphoryl Transfer Potential d. Voltage potential e. Oxidation strength of the synthase

Answers

The energy required to form ATP in the ATP Synthase is provided by Proton Flow and Voltage potential. These two factors cause a conformational change in the structure of ATP Synthase, which results in the formation of ATP molecules (option a and d).

ATP Synthase is an enzyme complex that converts ADP to ATP. The energy required for the formation of ATP is obtained from the electron transport chain and oxidative phosphorylation. The proton gradient that is established in the inner mitochondrial membrane during the electron transport chain is used to synthesize ATP through ATP Synthase.

The process is known as chemiosmotic coupling and it is the key mechanism behind ATP production in the cell. During the chemiosmotic coupling, protons (H+) are pumped out of the mitochondrial matrix into the intermembrane space. This results in the establishment of a proton gradient across the inner mitochondrial membrane.As the protons move back into the matrix through the ATP Synthase, the energy generated is used to produce ATP. This process is called oxidative phosphorylation and it is a crucial step in cellular respiration. Hence, options (a) and (d) are correct.

To know more about energy:

https://brainly.com/question/8630757

#SPJ11

Stroke volume is directly proportional to O preload O EDV and contractility. O contractility. O total peripheral resistance.

Answers

Stroke volume is directly proportional to preload (EDV) and contractility. These are two of the most important determinants of stroke volume. Total peripheral resistance does not have a direct effect on stroke volume.

What is stroke volume?

The volume of blood pumped out by the heart with each heartbeat is known as stroke volume. The ventricles eject a fixed volume of blood with each contraction, which is known as the stroke volume. The amount of blood pumped by the left ventricle into the aorta and by the right ventricle into the pulmonary artery is referred to as the stroke volume.

The three primary factors that influence stroke volume are preload, contractility, and afterload.

Preload: Preload is the volume of blood in the ventricles at the end of diastole (the relaxation phase of the cardiac cycle) before contraction. During diastole, the ventricles fill with blood. The more the ventricles are filled with blood, the more stretch they experience. The stretch on the heart muscle fibers is proportional to the quantity of blood in the ventricles. The greater the stretch, the greater the force of the contraction. As a result, increased preload stretches the ventricular walls, resulting in increased force of contraction and a greater stroke volume.

Contractility: Contractility refers to the strength of the heart's contractions. A healthy heart has a strong contractile force. The amount of blood pumped out of the heart is influenced by the force of the contraction. When the contractility of the heart increases, the heart beats with more force, resulting in an increase in stroke volume. When the contractility of the heart decreases, the heart beats with less force, resulting in a decrease in stroke volume.

Afterload: The resistance in the blood vessels that the heart must overcome to pump blood into the circulatory system is known as afterload. The resistance that the ventricle faces as it ejects blood into the arteries is referred to as afterload. Afterload can be affected by total peripheral resistance (TPR), which is the sum of all the peripheral resistances in the circulation. Since an increase in peripheral resistance raises afterload, it also reduces stroke volume.

Thus, the correct option is preload (EDV) and contractility.

To know more about Stroke volume, refer to the link below:

https://brainly.com/question/31667573#

#SPJ11

An antibiotic assay was conducted to determine if MH1 is resistant to the antibiotics Vancomycin (Van), Carbenicillin (Carb), and Gentamicin (Gen). In which of the following plates will you observe bacterial growth, IF MH1 is resistant to the antibiotics Vancomycin (Van) and Gentamicin (Gen). Note: This is a hypothetical scenario meant to help you with results interpretation. The results from your section's experiment might be different from what is described in this question.
a. LB only b. LB + Van c. LB + G d. LB + Carb

Answers

If MH1 is resistant to Vancomycin (Van) and Gentamicin (Gen), bacterial growth will be observed in the following plates:

a. LB only: In this plate, MH1 will grow since it is not sensitive to Vancomycin or Gentamicin. The absence of antibiotics allows the bacteria to thrive.

b. LB + Van: MH1 will grow in this plate as well since it is resistant to Vancomycin. The presence of Vancomycin will not inhibit its growth.

c. LB + G: MH1 will grow in this plate too as it is resistant to Gentamicin. The presence of Gentamicin will not hinder its growth.

d. LB + Carb: In this plate, bacterial growth will not be observed if MH1 is resistant to Carbenicillin. Carbenicillin is not mentioned as an antibiotic to which MH1 is resistant, so it may inhibit the growth of MH1 in this plate.

Therefore, the correct answer is d. LB + Carb.

To know more about  Carbenicillin visit:

https://brainly.com/question/30619014

#SPJ11

Other Questions
A person suffering from hyponatremia has a sodium ionconcentration in the blood of 0.119 MM and a total blood volume of5.0 LL .Part AWhat mass of sodium chloride would need to be added to the bloo Microbial adhesins can be found in which location? Choose allthat apply.in biofilmson bacterial ribosomeson host cellson bacterial pili and capsuleson cells at the portal of entry write an essay about your carrer path as an accountant Which of the following is NOT a broad ecosystem category? a. Low salt content, low biodiversity but minimum seasonality b. Areas of low salt content c. Many fluctuations based on seasonality d. High levels of biodiversity and salt content The swordtail crickets of the Hawaiian islands exemplify: O the influence of the formation of underlying hotspots on speciation, with crickets moving east to west over millions of years O strong sexual selection based upon courtship songs O occupation effects of different climactic zones/niches of islands O the evolutionary driving force of a shift to new food resources 1. Nutrients and oxygen for deep water animals comesfrom surface watersTrue or False2. reef corals are considered polypstrue or false3. Parapodia, in polychaete worms, are used for gasexchange and locomotiontrue or false W Edwards Deming came up with the "4 Absolutes." of TQM Select one: True False On the pGLO plasmid, what is the bla gene for? Group of answer choices It is the origin of replication so the bacterial cell can copy the plasmid. It codes for the green fluorescent protein. It allows us to select for bacterial cells that picked up the plasmid. It allows us to control whether the GFP gene is expressed or not. Which of the following transformations represent an increase in the entropy of the system. Choose all that apply 012 g C5H12 (gas, 309K) 12 g C5H12 (liquid, 309K) 4 mol CO (15.9 L, 212K) 4 mol CO how low-range hydrostatic pressure can be use toto destroy bacterial spores in food when combined with other antibacterial treatment. A Labrador breeder analyzed the pedigrees of two of her dogs and determined that the black male has a 25% chance of having the genotype BBEe and a 75% chance of having the genotype BbEe. Her yellow female has a 25% chance of having the genotype BBee and a 75% chance of having the genotype Bbee. Answer the following questions: a. Coat color in Labradors exhibits what genetic concept? Define this concept. b. What are all the possible genotypes for chocolate Labradors? A message signal m(t)=2sin( 1000 t) is used to modulate a 10 MHz carrier to generate a frequency modulated signal with frequency deviation delta f = 3kHz and waveform amplitude A=1 volt. Determine the following:a) The time domain expression of the modulated signal.b) The frequency deviation constant kf.c) The bandwidth estimate using Carsons rule. Brimco Company manufactures infant car seats for export in the South East Asia region. The price-demand equation and the monthly cost function for the production of x infant car seat as given, respectively, by: x=900030pC(x)=150000+30xwhere x is the number of infant car seats that can be sold at a price of p and C(x) is the total cost (in dollars) of producing x infant car seats. a. Find the profit function. b. How many infant car seats should the company manufacture each month to maximize its profit? What is the maximum monthly profit? How much should the company charge for each infant car seat? Suppose your utility function is given as U(x,y)=10x ^0.5 y^0.5Your income is equal to $100. One unit of x costs $10 and one unit of y costs $5. a) Using the Lagrange method, calculate the utility maximizing quantities of x and y. Also, calculate the Lagrange multiplier. Calculate the overall utility at this point. b) Now assume that good y gets taxed so that its price rises from $5 to $10. The price of x is still the same. Calculate the new optimal quantities consumed and the new utility level. In Type 1 diabetes the pancreas cannot produce enough insulin whereas in Type 2 diabetes the body cells become less responsive to insulin over time. True False if you encounter a grizzly bear while hiking, and you suddenly remember that you should first wave your arms and shout, this would be an example of information being brought into: In your own words explain at what ratio of input/naturalfrequencies system will have resonancePlease include as much information and as detailed as possible. Iwill upvote thank you so much! When you divide x^9 - 2 by the quantity of x minus the cube root3, the remainder is?a. 27b. 23c. 29d. 25 Assume the banking system contains the Jollowing amounts. Use this information to answer five questions. Instructions: Enter your responses rounded to the nearest whole number. a. Are the banks fully My black, heavy-coated dog is sitting on the grass in the sun on a hot day, panting. My friends white dog is with her, also sitting on the grass and panting. Compare and contrast the various ways in which the two dogs are gaining and losing heat (being careful to use the correct terminology). (10 marks)